A single frictionless roller-coaster car of mass m = 825 kg tops the first hill with speed v0 = 17.0 m/s at height h = 42.0 m. What is the speed of the car at (a) point A, (b) point B, and (c) point C? (d) How high will the car go on the last hill, which is too high for it to cross? (e) If we substitute a second car with twice the mass, what then are the answers to (a) through (d)?

Incomplete question check attachment for diagram of the question

Explanation:

Mass of car is m= 825kg

Initial velocity at first hill Vo=17m/s.

The initial velocity is in the direction then, Vx =17m/s

Height of hill is h=42m

a. At point, that is the maximum height and at maximum height for a projectile, it has only horizontal component of the velocity. So, it vertical component of the velocity is zero

Vy= 0m/s

Then, the magnitude of the velocity at A is

V= √Vox²+ Voy²

V=√17²+0²

V=√17²

V= 17m/s

b. At point B. Height is h/2,∴

half of initial PE has got converted into KE of the car

Using conservation of energy

K.E = P.E

½m•Vb² = mg(h/2)

m•Vb²= mgh

Divide both sides by m

Vb² =gh

Vb=√gh

Vb =√9.81×42

Vb=√470.88

Vb=21.7m/s

This is the velocity in the y-direction

V=√Vx²+Vy²

V=√17²+21.7²

V=√759.88

V=27.57m/s

c. Velocity at point C down the hill

At Point C. Here all the initial PE has got converted into KE of the car.

K.E = P.E

½m•Vc² = mgh

mVc² =2mgh

Divide both sides by m.

Vc² =2gh

Vc=√2gh

Vc= √2×9.81×43

Vc = √843.66

Vc = 29.05m/s

Also this is the y component of the velocity

Then, total velocity is

V=√Vx² +Vy²

V=√17²+29.05²

V=√1132.66

V=33.66m/s

d. At the last hill. Car can go only up to initial height h=42m

Using conservation of energy

∆K.E =∆ P.E

½mVf² -½mVi² = mgH - mgh

The initial velocity from the start is

Vi=17m/s

the final velocity is Vf=33.66m/s

The initial height is 0 h=0, from the bottom at point C.

Then, notice mass (m) is common to all the equation let divide through by m, so we have

½Vf²-½Vi² = gH

½ × 33.66² - ½ × 17² = 9.81H

566.498 - 144.5 =9.81H

421.998 =9.81H

H=421.998/9.81

H=43.017

H≈43m

Which is equal to the original height where the car incoming from.

e. If we double the mass of the car, it won't have effect on any of the answers, because all the answers are independent on mass, so, the answer still remains

1. At point A, V=17m/s

2. At point B, V =27.57m/s

3. At point C, V=33.66m/s

4. The car will still go as high as 43m as the previous answer

Note: the only thing that can make the mass to have effect is friction and air resistance and since they are neglected, then the answers are independent on mass

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting